3x-y=4
2x-3y=-9
To solve this system of equations, we can use a method called substitution or elimination.
Let's start with the substitution method:
Step 1: Solve one equation for one variable in terms of the other variable.
From the first equation, we can express y in terms of x:
3x - y = 4
y = 3x - 4
Step 2: Substitute the expression obtained in Step 1 into the other equation.
Substitute y = 3x - 4 into the second equation:
2x - 3(3x - 4) = -9
Step 3: Simplify and solve for x.
2x - 9x + 12 = -9
-7x + 12 = -9
-7x = -9 - 12
-7x = -21
x = -21 / -7
x = 3
Step 4: Substitute the value of x back into one of the original equations to solve for y.
Let's use the first equation:
3(3) - y = 4
9 - y = 4
-y = 4 - 9
-y = -5
y = 5
So the solution to the system of equations is x = 3 and y = 5.
Alternatively, you could use the elimination method:
Step 1: Multiply the first equation by 3 and the second equation by 2 to make the coefficients of x in both equations the same.
3(3x - y) = 3(4)
2(2x - 3y) = 2(-9)
9x - 3y = 12
4x - 6y = -18
Step 2: Multiply the second equation by 3 and subtract it from the first equation to eliminate the variable x.
9x - 3y - (12x - 18y) = 12 - (-18)
9x - 3y - 12x + 18y = 12 + 18
-3x + 15y = 30
Step 3: Solve the resulting equation for y.
-3x + 15y = 30
15y = 3x + 30
y = (3x + 30) / 15
y = (x + 10) / 5
Step 4: Substitute the value of y into one of the original equations to solve for x.
Let's use the first equation:
3x - ((x + 10) / 5) = 4
15x - (x + 10) = 20
15x - x - 10 = 20
14x - 10 = 20
14x = 20 + 10
14x = 30
x = 30 / 14
x = 15 / 7
Step 5: Substitute the value of x back into one of the original equations to solve for y.
Let's use the second equation:
2((15/7)) - 3y = -9
(30/7) - 3y = -9
-3y = -9 - (30/7)
-3y = (-63 - 30) / 7
-3y = -93 / 7
y = (-93/7) / -3
y = 31/7
So the solution to the system of equations is x = 15/7 and y = 31/7.